Submission #556945

#	Time	Username	Problem	Language	Result	Execution time	Memory
556945		DanShaders	Misspelling (JOI22_misspelling)	C++17	28 / 100	4065 ms	124928 KiB

misspelling

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;

namespace x = __gnu_pbds;
template <typename T>
using ordered_set = x::tree<T, x::null_type, less<T>, x::rb_tree_tag, x::tree_order_statistics_node_update>;

template <typename T>
using normal_queue = priority_queue<T, vector<T>, greater<>>;

#define all(x) begin(x), end(x)
#define sz(x) ((int) (x).size())
#define x first
#define y second
using ll = long long;
using ld = long double;

#define int ll

const int N = 1 << 19, C = 26, MOD = 1e9 + 7;

ll dp[N][C];

int t[2][2 * N];

signed main() {
	cin.tie(0)->sync_with_stdio(0);
	int n, m;
	cin >> n >> m;
	fill(t[0], t[2], -1);
	for (int i = 0; i < m; ++i) {
		int u, v;
		cin >> u >> v;
		--u, --v;
		int l = N + min(u, v) + 1, r = N + max(u, v);
		bool at = u > v;
		while (l <= r) {
			if (l & 1)	t[at][l] = max(t[at][l], min(u, v));
			if (!(r & 1))	t[at][r] = max(t[at][r], min(u, v));
			l = (l + 1) / 2;
			r = (r - 1) / 2;
		}
	}
	for (int i = 0; i < 2; ++i) {
		for (int j = 0; j < N; ++j) {
			t[i][2 * j] = max(t[i][2 * j], t[i][j]);
			t[i][2 * j + 1] = max(t[i][2 * j + 1], t[i][j]);
		}
	}

	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < C; ++j) {
			dp[i][j] = 1;
		}
	}

	for (int i = n - 1; i > 0; --i) {
		for (int c0 = 0; c0 < C; ++c0) {
			int j0 = t[0][N + i], j1 = t[1][N + i];
			for (int j = max(j0, j1) + 1; j < i; ++j) {
				for (int c1 = 0; c1 < C; ++c1) {
					if (c0 != c1) {
						(dp[j][c1] += dp[i][c0]) %= MOD;
					}
				}
			}
			if (j0 < j1) {
				for (int j = min(j0, j1) + 1; j <= max(j0, j1); ++j) {
					for (int c1 = 0; c1 < c0; ++c1) {
						(dp[j][c1] += dp[i][c0]) %= MOD;
					}
				}
			} else {
				for (int j = min(j0, j1) + 1; j <= max(j0, j1); ++j) {
					for (int c1 = c0 + 1; c1 < C; ++c1) {
						(dp[j][c1] += dp[i][c0]) %= MOD;
					}
				}
			}
		}
	}
	cout << accumulate(dp[0], dp[1], 0ll) % MOD;
}

• Subtask 1: 8 / 8
• Subtask 2: 20 / 20
• Subtask 3: 0 / 29
• Subtask 4: 0 / 32
• Subtask 5: 0 / 11